Light - Reflection and Refraction Class 10 MCQs for Science CBSE Board

Light plays a crucial role in our everyday life, from seeing objects around us to understanding how mirrors, lenses, cameras, and spectacles work. In this chapter, students explore important concepts such as reflection of light, refraction of light, spherical mirrors, lenses, image formation, refractive index, and the applications of these principles in real-world situations. Practicing these Light - Reflection and Refraction Class 10 MCQs is one of the most effective ways to strengthen conceptual understanding and prepare for CBSE Board examinations.

This collection of Class 10 Science MCQs is designed to help students revise key concepts quickly, test their knowledge, and improve accuracy in objective-type questions. Whether you are preparing for school tests, periodic assessments, or board exams, these questions cover the most important topics from Chapter 10 in a simple and exam-focused format. For broader preparation, students can also explore our complete collection of Class 10 MCQs and chapter-wise MCQs to practice all subjects and improve overall exam performance.

Each question is created to support quick revision, concept clarity, and competency-based learning. Along with answers and explanations, these MCQs help students understand not just the correct option but also the reasoning behind it, making preparation more effective for both exams and future learning.

Practice Light - Reflection and Refraction Class 10 MCQs

Before attempting the questions, make sure you have revised the basic concepts of reflection, refraction, spherical mirrors, lenses, image formation, and related formulas. This will help you answer questions more confidently and improve your accuracy.

Q. A current of 0.75 A is drawn by a filament of an electric bulb for 12 minutes. Find the amount of electric charge that flows through the circuit.

(A) 9 C
(B) 90 C
(C) 540 C
(D) 720 C

Answer: C

Explanation:

Given: Current (I) = 0.75 A, Time (t) = 12 minutes = 12 × 60 seconds = 720 s.
Using Formula: Q = I × t = 0.75 × 720 = 540 C.

Q. Which of the following correctly defines the direction of conventional electric current?

(A) In the direction of flow of electrons.
(B) From the negative terminal to the positive terminal of a cell.
(C) In the direction of flow of positive charges.
(D) It is randomly oriented depending on the conductor type.

Answer: C

Explanation:

By historical convention, the direction of electric current is considered to be the direction of flow of positive charges, which moves from the positive terminal to the negative terminal. This is exactly opposite to the physical drift direction of free electrons.

Q. How much work is done in moving a charge of 3 C across two points having a potential difference of 15 V?

(A) 5 J
(B) 45 J
(C) 0.2 J
(D) 30 J

Answer: B

Explanation:

Given: Charge (Q) = 3 C, Potential Difference (V) = 15 V.
Using Formula: Work Done (W) = V × Q = 15 × 3 = 45 Joules.

Q. The physical quantity that remains constant when resistors are connected in a series circuit is:

(A) Potential difference across each resistor
(B) Electric current through each resistor
(C) Both current and potential difference
(D) Total electric resistance

Answer: B

Explanation:

In a series circuit configuration, there is only one continuous path for charge carriers to flow. Therefore, the volume of electric current passing through each resistor remains identical, whereas the voltage drops divide across individual loads.

Q. If the length of a metallic wire is doubled while its cross-sectional area is halved, its new resistance will become:

(A) Two times
(B) Four times
(C) Half
(D) Unchanged

Answer: B

Explanation:

Resistance is given by R = ρ(L/A). If the new length L' = 2L and new cross-sectional area A' = A/2, then the new resistance R' = ρ(2L / (A/2)) = 4 × ρ(L/A) = 4R. Hence, total loop resistance scales by four times.

Q. An electrical appliance has a resistance of 44 ohms and operates on a 220 V supply line. What is the current drawn by it?

(A) 5 A
(B) 0.2 A
(C) 10 A
(D) 2.5 A

Answer: A

Explanation:

Given: Resistance (R) = 44 Ω, Voltage (V) = 220 V.
By Ohm's Law: I = V / R = 220 / 44 = 5 A.

Q. The resistivity of a certain material wire does not depend on which of the following parameters?

(A) Nature of the material
(B) Temperature of the conductor
(C) Length of the wire
(D) All of the above affect resistivity

Answer: C

Explanation:

Resistivity (ρ) is an intrinsic micro-property of a material. It changes only with the material's structural nature and operating temperature. It is entirely independent of the macroscopic dimensions like the length or cross-sectional area of the wire sample.

Q. Three resistors of resistances 2 Ω, 4 Ω, and 6 Ω are connected in parallel. The equivalent resistance of the combination will be:

(A) Greater than 6 Ω
(B) Exactly 12 Ω
(C) Less than 2 Ω
(D) Between 4 Ω and 6 Ω

Answer: C

Explanation:

In a parallel network, the reciprocal equivalent resistance formula (1/Rp = 1/R1 + 1/R2 + ...) guarantees that the net total resistance will always drop below the value of the smallest single individual branch resistor in the loop. Since the smallest value here is 2 Ω, the total network resistance must be less than 2 Ω.

Q. Which device is used to measure electric potential difference in a circuit and how is it connected?

(A) Ammeter, connected in series
(B) Voltmeter, connected in parallel
(C) Ammeter, connected in parallel
(D) Voltmeter, connected in series

Answer: B

Explanation:

A voltmeter measures the electric potential difference across two unique circuit nodes. It is structurally built with an exceptionally high internal impedance and must always be placed in parallel with the targeted device to avoid drawing significant bypass line current.

Q. According to Joule's heating law, the heat produced in a resistor is:

(A) Inversely proportional to the square of the current
(B) Directly proportional to the square of the current
(C) Inversely proportional to the resistance
(D) Directly proportional to the square root of time

Answer: B

Explanation:
Joule's Law of heating states that H = I²Rt. This clearly asserts that heat generated in a standard conductor varies directly as the square of the electric current (I²) flowing through that pathway.

Q. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:

(A) 100 W
(B) 75 W
(C) 50 W
(D) 25 W

Answer: D

Explanation:

Resistance of the bulb filament is R = V² / P = (220)² / 100 = 484 Ω. When operated at the newly reduced voltage V' = 110 V, the actual power consumed drops to: P' = (V')² / R = (110)² / 484 = 12,100 / 484 = 25 W.

Q. The commercial unit of electric energy is kilowatt-hour (kWh). 1 kWh is equal to:

(A) 3.6 × 10⁵ J
(B) 3.6 × 10⁶ J
(C) 0.36 × 10⁶ J
(D) 36 × 10⁶ J

Answer: B

Explanation:

1 kWh = 1 kilowatt × 1 hour = 1000 Watts × 3600 seconds = 3,600,000 Joules = 3.6 × 10⁶ J.

Q. Why are the filaments of incandescent electric lamps generally made of tungsten?

(A) It has a very low melting point.
(B) It reacts readily with atmospheric oxygen at high temperatures.
(C) It has a very high melting point and high resistivity.
(D) It is a poor conductor of electricity at room temperature.

Answer: C

Explanation:

Tungsten is selectively preferred for standard incandescent lightbulbs because it possesses an extraordinarily high melting point (approx 3422°C) combined with high operational thermal stability, preventing filament collapse or rapid oxidation at white-hot luminous temperatures.

Q. A cylindrical conductor of length 'L' and uniform area of cross-section 'A' has a resistance 'R'. Another conductor of length '2L' and resistance 'R' of the same material has an area of cross-section equal to:

(A) A/2
(B) 2A
(C) 4A
(D) A

Answer: B

Explanation:

Resistance R = ρL/A. For a secondary wire of the same material (so ρ matches) with double length (2L) to hold the identical absolute resistance R, we balance the relation: R = ρ(2L)/A'. Equating both formulas gives ρL/A = ρ(2L)/A', which maps directly to A' = 2A.

Q. What happens to the total circuit current when more resistors are added in a parallel combination across a fixed voltage source?

(A) The current increases.
(B) The current decreases.
(C) The current remains exactly the same.
(D) The current fluctuates unpredictably.

Answer: A

Explanation:

Adding extra alternative pathways in a parallel setup decreases the overall net equivalent resistance of the active circuit network. Since net resistance drops while the main driving source voltage remains fixed, the primary current drawn directly from the battery must increase via Ohm's law (I = V/R).

Q. If 10²⁰ electrons pass through a given cross-section of a wire in 4 seconds, what is the current flowing through it? (Charge of an electron = 1.6 × 10^-19 C)

(A) 4 A
(B) 16 A
(C) 2.5 A
(D) 40 A

Answer: A

Explanation:

Total net charge Q = n × e = 10²⁰ × 1.6 × 10−19 C = 16 C.
Current (I) = Q / t = 16 C / 4 s = 4 A.

Q. An electrical fuse works on the principle of:

(A) Chemical effect of electric current
(B) Magnetic effect of electric current
(C) Heating effect of electric current
(D) Optical effect of electric current

Answer: C

Explanation:

A safety circuit fuse incorporates a specialized low-melting-point metallic ribbon. When internal current loads spike past safe engineering thresholds, rapid internal thermal energy builds via the heating effect of electric current (Joule's heating), melting the element to cleanly split the loop open.

Q. Two copper wires A and B have lengths in the ratio 1:2 and areas of cross-section in the ratio 2:1. The ratio of their resistivities is:

(A) 1:1
(B) 1:4
(C) 4:1
(D) 1:2

Answer: A

Explanation:

Resistivity is an intrinsic macro-independent constant dependent exclusively on the structural chemical material matrix and thermal environment. Since both wires A and B are explicitly stated to be copper, their resistivity coefficients are identical, preserving a 1:1 ratio.

Q. Ohm's law is valid only when the ________ of the conductor remains constant.

(A) Current
(B) Potential difference
(C) Temperature
(D) Resistance

Answer: C

Explanation:

Ohm's Law (V = IR) holds structurally accurate only under static experimental parameters where environmental and physical variables—specifically the physical temperature of the metallic conductor matrix—are strictly kept uniform during logging.

Q. A wire of resistance R is cut into five equal pieces. These pieces are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is:

(A) 1/25
(B) 5
(C) 1/5
(D) 25

Answer: D

Explanation:

When a standard wire of resistance R is cleanly sectioned into 5 identical pieces, each piece holds an individual resistance value of r = R/5. When these 5 individual segments are linked across a parallel mesh: 1/R' = 5 × (1 / r) = 5 × (5 / R) = 25/R. Rearranging this relationship yields the exact ratio: R / R' = 25.

Q. The V-I graph for a certain metallic wire at two different temperatures T1 and T2 is given. If the straight line for T1 is steeper (closer to the V-axis) than T2, which temperature is higher? (Assume V is on the y-axis and I is on the x-axis)

(A) T1 > T2
(B) T2 > T1
(C) T1 = T2
(D) Cannot be determined from the graph

Answer: A

Explanation:

On a V-I graph where Potential Difference (V) tracks on the y-axis, the instantaneous line slope directly represents the Resistance value (R = V/I). A steeper slope marks a higher resistance. Because the electrical resistance of pure metallic conductors scales upward with temperature, the higher resistance curve (T1) signals a higher corresponding thermal state (T1 > T2).

Q. An electric iron draws a current of 5 A when connected to a 220 V line. Calculate the heat developed in 30 seconds.

(A) 33,000 J
(B) 1,100 J
(C) 6,600 J
(D) 2,200 J

Answer: A

Explanation:
Given: Current (I) = 5 A, Voltage (V) = 220 V, Time (t) = 30 seconds.
Using formula: Heat (H) = V × I × t = 220 × 5 × 30 = 33,000 Joules.

Q. What is the maximum resistance that can be made using five resistors each of value 1/5 Ω?

(A) 1/5 Ω
(B) 1 Ω
(C) 5 Ω
(D) 25 Ω

Answer: B

Explanation:
Maximum prospective resistance is achieved when elements are linked in a strict series combination. Therefore, R_max = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 5/5 = 1 Ω.

Q. What is the minimum resistance that can be made using five resistors each of value 1/5 Ω?

(A) 1/25 Ω
(B) 1/5 Ω
(C) 1 Ω
(D) 1/125 Ω

Answer: A

Explanation:
Minimum prospective network resistance is created when all component pieces are wired in a fully parallel configuration: 1/Rmin = 5 × (1 / (1/5)) = 5 × 5 = 25. Inverting this reciprocal structural term gives Rmin = 1/25 Ω.

Q. Which of the following variables is plotted on a graph to verify Ohm's law, and what is the shape of the resulting curve for an ohmic conductor?

(A) V vs I, Parabolic curve
(B) V vs I, Straight line passing through the origin
(C) P vs I, Straight line passing through the origin
(D) V vs R, Exponential curve

Answer: B

Explanation:
Ohm's fundamental law establishes that V is directly proportional to I under standard limits. Tracking Potential Difference (V) against line Current (I) generates a clean linear straight-line relationship tracking precisely through the grid origin coordinates (0,0).

Q. A current of 2 A flows through a 10 Ω resistor for 1 minute. The electrical energy consumed is:

(A) 240 J
(B) 2400 J
(C) 40 J
(D) 400 J

Answer: B

Explanation:
Given: I = 2 A, R = 10 Ω, t = 1 minute = 60 seconds.
Energy = I² × R × t = (2)² × 10 × 60 = 4 × 10 × 60 = 2400 Joules.

Q. Which of the following substances has the lowest electrical resistivity at room temperature?

(A) Constantan
(B) Glass
(C) Silver
(D) Silicon

Answer: C

Explanation:
Silver is universally recognized as holding the premier electrical conductivity index among common crystalline metals at standard room temperatures, boasting the absolute lowest resistivity values (~1.59 × 10−8 Ω·m) to limit path losses.

Q. Two electric bulbs have specifications 220 V, 60 W and 220 V, 100 W. The filament of which bulb will have a higher resistance value?

(A) The 100 W bulb
(B) The 60 W bulb
(C) Both will have equal resistance since voltage rating is identical
(D) It depends on the current passing through the circuit

Answer: B

Explanation:

Using the physical relationship R = V² / P, resistance values map inversely to wattage ratings when active operational voltage thresholds match. Therefore, the lower-wattage lighting appliance (60 W) will possess a higher absolute warm filament resistance than the 100 W element.

Q. When a variable resistor (rheostat) is connected in a circuit, its main purpose is to:

(A) Change the voltage without altering the current
(B) Change the current without altering the source voltage
(C) Measure the exact resistance value of the load
(D) Safeguard the circuit against ground faults

Answer: B

Explanation:

A variable inline rheostat regulates dynamic circuit parameters by shifting its active physical internal path length to alter current loads smoothly without changing the underlying fixed main source feed voltage.

Q. The charge flowing through a circuit per unit time is called:

(A) Conductance
(B) Potential energy
(C) Electric current
(D) Resistivity

Answer: C

Explanation:

By strict definition, the volumetric rate at which an active electrical charge stream drifts past an observed structural cross-sectional line reference per unit time interval is classified as electric current (I = Q/t).

Quick Chapter Overview

Chapter 10 focuses on the behavior of light when it reflects from surfaces and refracts while passing through different media. Students learn the laws governing reflection and refraction, image formation by mirrors and lenses, sign conventions, lens formula, mirror formula, magnification, and practical applications of optical devices.

This chapter combines both conceptual understanding and numerical problem-solving. A strong grasp of formulas and ray diagrams is essential for scoring well.

Key Areas at a Glance

Important Concepts Covered in These MCQs

The MCQs on this page are designed to cover the major learning outcomes of the chapter.

Reflection Concepts

• Reflection of light
• Laws of reflection
• Regular reflection
• Diffused reflection
• Incident ray and reflected ray
• Angle of incidence
• Angle of reflection

Spherical Mirrors

• Concave mirror
• Convex mirror
• Principal focus
• Radius of curvature
• Pole
• Principal axis
• Mirror formula
• Magnification

Refraction Concepts

• Refraction of light
• Optical density
• Refractive index
• Speed of light in different media
• Snell's law
• Refraction through glass slab

Lens Concepts

• Convex lens
• Concave lens
• Optical centre
• Principal focus
• Lens formula
• Magnification by lens
• Power of lens

Practical Applications

• Rear-view mirrors
• Magnifying glasses
• Cameras
• Microscopes
• Spectacles
• Optical instruments

Quick Revision Notes

Use these points for a fast chapter recap before attempting MCQs.

Reflection

• Angle of incidence = Angle of reflection.
• Incident ray, reflected ray, and normal lie in the same plane.
• Reflection occurs when light strikes a reflecting surface.

Refraction

• Refraction is the bending of light when light enters another medium.
• Light bends towards the normal when moving from a rarer medium to a denser medium.
• Light bends away from the normal when moving from a denser medium to a rarer medium.

Mirror Formula

1/f = 1/v + 1/u

Where:

• f = focal length
• v = image distance
• u = object distance

Lens Formula

1/f = 1/v - 1/u

Where:

• f = focal length
• v = image distance
• u = object distance

Magnification

For Mirrors:

m = -v/u = hᵢ/hₒ

For Lenses:

m = v/u = hᵢ/hₒ

Where:

• hᵢ = height of image
• hₒ = height of object

Power of Lens

P = 1/f (when f is in metres)

OR

P = 100/f (when f is in centimetres)

Unit: Dioptre (D)

Tips to Score Better in Reflection and Refraction MCQs

Learn Sign Conventions Properly

Many students know the formulas but lose marks because they apply incorrect signs. Always revise mirror and lens sign conventions carefully.

Practice Ray Diagrams

Image formation questions often become easier when you visualize ray paths instead of memorizing answers.

Memorize Core Formulas

Keep mirror formula, lens formula, magnification formula, and power of lens formula at your fingertips.

Focus on Concept-Based Questions

Recent CBSE papers increasingly test understanding rather than direct memorization.

Solve Numerical Questions Regularly

A few minutes of daily practice can significantly improve speed and accuracy.

Common Mistakes Students Should Avoid

Mixing Up Concave and Convex Properties

Many students confuse image characteristics formed by concave mirrors and convex mirrors.

Ignoring Sign Convention

Incorrect sign usage often leads to wrong numerical answers despite correct calculations.

Memorizing Without Understanding

Conceptual clarity is more useful than remembering isolated facts.

Skipping Ray Diagrams

Ray diagrams help verify image formation and improve conceptual understanding.

Forgetting Units

Always check units while solving numerical questions related to focal length and power.

Important Exam Focus Areas

If you are revising before a test or board examination, pay special attention to these topics.

Conclusion

These Light - Reflection and Refraction Class 10 MCQs are designed to help students revise important concepts, evaluate their understanding, and prepare effectively for CBSE examinations. Since this chapter combines theory, formulas, ray diagrams, and numerical applications, consistent practice is the key to mastering it.

Before attempting the questions, revise the important concepts and formulas discussed above. Once you complete the MCQs, review the explanations carefully to strengthen your understanding and improve your performance in future tests and board examinations.